y1=x,y2=x^2,y3=e^x為方程y''+p(x)y'+q(x)y=f(x)的三個特解這個二階微分方程顯然有兩個通解,那么顯然x^2-x和e^x -x就是y''+p(x)y'+q(x)y=0的通解,于是y''+p(x)y'+q(x)y=f(x)的通解就是y=A*(x^2-x) +B*(e^x -x) + x,A...錯了錯了……y1=x^2y1=x^2 ？那y2等于什么呢？把題目寫清楚點啊微分方程通解和特解，已知y1=x^2，y2=e^x，y3=e^-x為方程y''+p(x)y'+p(x)y=f(x)的三個特解，求通解。我錯了……微分方程通解和特解，已知y1=x^2，y2=e^x，y3=e^-x為方程y''+p(x)y'+p(x)y=f(x)的三個特解，求通解。我錯了……還是同樣的做法你就想著非齊次方程兩個特解的差就是對應齊次方程的通解y1=x^2，y2=e^x，y3=e^-x為方程y''+p(x)y'+q(x)y=f(x)的三個特解這個二階微分方程顯然有兩個通解，那么顯然x^2-e^x和x^2 -e^-x 就是y''+p(x)y'+q(x)y=0的通解，于是y''+p(x)y'+q(x)y=f(x)的通解就是y=A*(x^2-e^x) +B*(x^2 -e^-x) + x^2，AB為常數(shù)